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2 years ago

Describe an electron cloud. Then, name and explain a particular atomic model that used the concept of the electron cloud.

Chemistry

1 answer:

pochemuha2 years ago

3 0

Electron cloud is the region around the nucleus in an atom where we can locate an electron.

The concept of electron cloud model was introduced by the Schrodinger and Heisenberg. According to this model, it would be difficult to know the position of the electrons in an atom and they are not particles that orbit around the nucleus. We can only expect the electrons to be present in specific areas called the electron clouds around the nucleus. It is the quantum mechanical model that used the concept of electron clouds. According to the model, the electron cloud or an orbital is a space around the nucleus in an atom where the probability of finding an electron is 90%. It explains that electrons show wave nature. It is difficult to determine the exact position and momentum of an electron in an atom.

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According to Newton's first law of physics, what will a bowling ball do as it rolls down the bowling alley?

Nitella [24]

Answer:

Because there is no friction, Newton's first law states that the ball should continue to roll. (continue at a constant speed)

Explanation:

Unless acted upon by an unbalanced force, Newton's first law of motion states that an object at rest stays at rest and an object in motion stays in motion with the same speed and direction.

6 0

2 years ago

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

Over-the-counter hydrogen peroxide solutions are 3% (w/v). What is this concentration in moles per liter?

Marina CMI [18]

<span>1 ml of water weighs 1 gram so 1 liter (1000 ml) weighs 1000 grams. A 3% solution (3% = 0.03) of hydrogen peroxide (w/v) would contain 1000 grams x 0.03 or 30 grams. The chemical formula of hydrogen peroxide is H2O2 and a mole weighs 34.0147 grams/mole. So 30 grams of H2O2 divided by 34.0147 grams/mole equals 0.88 moles of H2O2. The concentration of a 3% (w/v) hydrogen peroxide solution therefore contains 30 grams of H202 (or 0.88 moles of H202) per in a liter of water (or 1000 grams H20) would thus be 0.88 moles H2O2 per liter (0.88 moles H2O2/l) .</span>

5 0

2 years ago

Several different species of birds, including heron, flamingos, and skimmers, can all successfully feed along the coast because

alexandr402 [8]

The correct answer is resource partitioning.

The concept of resource partitioning is applicable in the branch of ecology. It signifies towards the procedure by which natural selection mediates competing species into distinct patterns of different niches or resource utilization.

When the species differentiate a niche to prevent competition for food resources, it is known as resource partitioning. At certain times, the competition is among the species, known as interspecific competition, and at sometimes it is among the individuals of the similar species, that is, intraspecific competition.

3 0

2 years ago

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

joja [24]

Answer:

6.24 x 10-3 M

Explanation:

Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

$Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}$

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change $x$ , we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

$5.2x10^{-5}=\frac{x*x}{0.749-x}$

Thus, we obtain a quadratic equation whose solution is:

$x_1=-0.00627M\\x_2=0.00624M$

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.

Best regards.

4 0

2 years ago

Read 2 more answers